Evaluate the following limits in Exercises 1 to 22: limₓ→₀ (sin ax + bx)/(ax + sin bx), a, b, a + b ≠ 0

Solution:

At x = 0, the value of the given function takes the form 0/0, which is an indeterminate form.

So we will evaluate the given limit differently.

The given limit is:

limₓ→₀  (sin ax + bx)/(ax + sin bx)

Multiply and divide sin ax by ax; and sin bx by bx,

= limₓ→₀ [ax (sin ax/ax) + bx] / [ax + bx (sin bx/bx)]

= [limₓ→₀ (sin ax/ax) x limₓ→₀ (ax) + limₓ→₀ (bx)] / [limₓ→₀ (ax) + limₓ→₀ (bx)(limₓ→₀ sin bx/bx)]

Now, as x → 0 ⇒ ax → 0 and bx → 0.

= [lim ₐₓ→₀ (sin ax/ax) x limₓ→₀ (ax) + limₓ→₀ (bx)] / [limₓ→₀ (ax) + limₓ→₀ (bx)(lim bₓ→₀ sin bx/bx)]

= [limₓ→₀ (ax) + limₓ→₀ (bx)]/[limₓ→₀ (ax) + limₓ→₀ (bx)] [∵ limₓ→₀ (sin x/x) = 1]

= (limₓ→₀ ax + bx)/(limₓ→₀ ax + bx)

= limₓ→₀ (1)

= 1

NCERT Solutions Class 11 Maths Chapter 13 Exercise 13.1 Question 20

Evaluate the following limits in Exercises 1 to 22: limₓ→₀ (sin ax + bx)/(ax + sin bx), a, b, a + b ≠ 0

Summary:

The value of the limit limₓ→₀ (sin ax + bx)/(ax + sin bx), a, b, a + b ≠ 0 is 1